= Exact tidal acceleration between separated masses
{title2=$\Delta a=2\mu L/[R^3(1-L^2/(4R^2))^2]$}
Stipulate a fixed source <point mass> $M=10^{12}\,\mathrm{kg}$, two equal <test particles> of <mass> $m=1\,\mathrm{kg}$, and $G=6.67430\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$. This is a chosen model, not a measurement of a particular object. Place the <test particles> on the outward radial axis at $R-L/2$ and $R+L/2$, with midpoint distance $R=100\,\mathrm m$ and separation $L=2\,\mathrm m$. Neglect their mutual attraction, other sources and all supporting <forces>.
Each signed radial <gravitational acceleration> is inward:
$$
a_-=-\frac{\mu}{(R-L/2)^2}\approx-0.0068098153\,\mathrm{m\,s^{-2}},\qquad a_+=-\frac{\mu}{(R+L/2)^2}\approx-0.0065427899\,\mathrm{m\,s^{-2}}.
$$
Their separation has outward relative <acceleration>
$$
\ddot L=a_+-a_-=\mu\left[\frac{1}{(R-L/2)^2}-\frac{1}{(R+L/2)^2}\right]=\frac{2\mu L}{R^3}\left(1-\frac{L^2}{4R^2}\right)^{-2}.
$$
The exact value is $2.6702540\times10^{-4}\,\mathrm{m\,s^{-2}}$. The midpoint <tidal tensor> predicts $2\mu L/R^3=2.6697200\times10^{-4}\,\mathrm{m\,s^{-2}}$, an underestimate of $0.019999\%$ relative to the exact value. For these equal $1\,\mathrm{kg}$ <masses>, the difference of external <forces> has the same numerical value in the <SI> unit of <force>, the <newton (unit)>. It is not an additional mutual <force> between the <test particles>.
The exact formula requires $L<2R$ and describes the instantaneous relative <acceleration>. Because the two sample <positions> are symmetric about the midpoint, the <Taylor expansion> error relative to the linear result starts at order $(L/R)^2$; a one-sided comparison with a reference point generally has relative error of order $L/R$. If either <test particle> enters an extended source, use that source's interior <Newtonian gravitational field> instead of extrapolating the singular <point mass> model.
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